Problem 40 - Chemistry Central Science

Question

An aerosol spray can with a volume of \(250 \mathrm{~mL}\) contains \(2.30 \mathrm{~g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(23^{\circ} \mathrm{C}\), what is the pressure in the can? (b) What volume would the propane occupy at STP? (c) The can says that exposure to temperatures above \(130^{\circ} \mathrm{F}\) may cause the can to burst. What is the pressure in the can at this temperature?

Step-by-Step Solution

Answer

(a) The pressure in the can at \(23^{\circ} \mathrm{C}\) is \(5.18 \mathrm{~atm}\). (b) The volume of propane at STP is \(1.12 \mathrm{~L}\). (c) The pressure in the can at \(130^{\circ} \mathrm{F}\) is \(6.89 \mathrm{~atm}\).

Step 1: Calculate the number of moles of propane (C3H8)

Since the mass of propane in the can is given, we need to calculate the number of moles of propane. To do this, we find the molar mass of propane and then divide the mass by the molar mass. Molar mass of C3H8 = \((3 * 12.01) + (8 * 1.01) = 44.11 \mathrm{~g/mol}\) Number of moles, \(n = \frac{2.30 \mathrm{~g}}{44.11 \mathrm{~g/mol}} = 0.0521 \mathrm{~mol}\)

Step 2: (a) Calculate the pressure in the can

Now, we can use the Ideal Gas Law to calculate the pressure in the can. The given temperature must be converted to Kelvin (K): \(T = 23^{\circ} \mathrm{C} + 273.15 = 296.15 \mathrm{~K}\) Volume of the can is given as \(250 \mathrm{~mL}\), which needs to be converted to liters: \(V = 250 \mathrm{~mL} = 0.250 \mathrm{~L}\) Using the Ideal Gas Law, \(PV = nRT\), we can solve for pressure \(P\): \(P = \frac{nRT}{V} = \frac{0.0521 \mathrm{~mol} * 0.0821 \mathrm{~L \cdot atm/mol \cdot K} * 296.15 \mathrm{~K}}{0.250 \mathrm{~L}} = 5.18 \mathrm{~atm}\) So, the pressure in the can is \(5.18 \mathrm{~atm}\).

Step 3: (b) Calculate the volume of propane at STP

The volume of the propane at STP can also be calculated using the Ideal Gas Law. Here, \(T = 273.15 \mathrm{~K}\) (0 °C) and the standard pressure is \(P = 1 \mathrm{~atm}\). Using the Ideal Gas Law, we can solve for the volume, \(V\): \(V = \frac{nRT}{P} = \frac{0.0521 \mathrm{~mol} * 0.0821 \mathrm{~L \cdot atm/mol \cdot K} * 273.15 \mathrm{~K}}{1 \mathrm{~atm}} = 1.12 \mathrm{~L}\) So, the volume of propane at STP is \(1.12 \mathrm{~L}\).

Step 4: (c) Calculate the pressure in the can at 130 °F

In this case, we need to convert the given temperature, \(130^{\circ} \mathrm{F}\), to Kelvin. To do that, first, convert °F to °C using the following formula: \(T_{°C} = (T_{°F} - 32) * 5 / 9 \). Then, convert °C to Kelvin (K) by adding 273.15. \(T = (130 - 32) * \frac{5}{9} \mathrm{~°C} = 54.44 \mathrm{~°C}\) \(T = 54.44 \mathrm{~°C} + 273.15 = 327.59 \mathrm{~K}\) Using the Ideal Gas Law (\(PV = nRT\)), we can solve for the pressure in the can at this temperature: \(P = \frac{nRT}{V} = \frac{0.0521 \mathrm{~mol} * 0.0821 \mathrm{~L \cdot atm/mol \cdot K} * 327.59 \mathrm{~K}}{0.250 \mathrm{~L}} = 6.89 \mathrm{~atm}\) So, the pressure in the can at \(130^{\circ} \mathrm{F}\) is \(6.89 \mathrm{~atm}\).